An object of length 4 cm is placed in front of a concave mirror at distance 30 cm. The focal length of mirror is 15 cm. a. Where will the image form? B. What will be the length of image?

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the given values
- Length of the object (h_o) = 4 cm
- Object distance (u) = -30 cm (negative due to sign convention)
- Focal length (f) = -15 cm (negative for concave mirror)

### Step 2: Use the mirror formula to find the image distance (v)
The mirror formula is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]

Substituting the values:
\[ \frac{1}{v} + \frac{1}{-30} = \frac{1}{-15} \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{30} \]

### Step 4: Find a common denominator
The common denominator for -15 and 30 is 30:
\[ \frac{1}{-15} = \frac{-2}{30} \]
So,
\[ \frac{1}{v} = \frac{-2}{30} + \frac{1}{30} = \frac{-2 + 1}{30} = \frac{-1}{30} \]

### Step 5: Solve for v
Taking the reciprocal gives:
\[ v = -30 \, \text{cm} \]

### Step 6: Interpret the result
The negative sign indicates that the image is formed in front of the mirror, at a distance of 30 cm.

### Step 7: Calculate the height of the image (h_i) using magnification
The magnification (M) is given by:
\[ M = \frac{h_i}{h_o} = -\frac{v}{u} \]

Substituting the known values:
\[ M = -\frac{-30}{-30} = -1 \]

### Step 8: Find the height of the image
Using the magnification:
\[ \frac{h_i}{4} = -1 \]

Cross-multiplying gives:
\[ h_i = -4 \, \text{cm} \]

### Step 9: Interpret the height of the image
The negative sign indicates that the image is inverted. Thus, the length of the image is 4 cm.

### Final Answers:
a. The image is formed at a distance of 30 cm in front of the mirror.
b. The length of the image is 4 cm (inverted).

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